Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT -Molecular weight: (MW) sum of the atomic weights of all the atoms in a molecule of the substance. -H 2 O = 18.0 amu (2 x 1.0 amu for two H atoms plus 16.0 amu from one O atom)

Formula weight (FW) – sum of the atomic weights of all atoms in a formula unit of a compound. Whether molecular or not. Sodium chloride, NaCl (formula unit) formula weight of amu (22.99 amu from Na plus amu from Cl). NaCl is ionic (no molecular weight)

3.2 THE MOLE CONCEPT -Developed to deal with the enormous numbers of molecules or ions in samples of substances. -Mole: the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12.

The number of atoms in a 12-g sample of carbon-12 is called Avogadro’s Number (N A ). = 6.02 x 10^23 A mole of a substance contains Avogadro’s number of molecules (6.02 x 10^23) Mole (much like dozen or gross) refers to a particular number of molecules. 1 mole of ethanol = 6.02 x 10^23 ethanol molecules When talking about ionic substances we are referring to the number of formula units not molecules.

One mole of sodium carbonate Na 2 CO 3 Contains: 6.02 x 10^23 Na 2 CO 3 units Which contains: 2 x 6.02 x 10^23 Na+ ions 1 x 6.02 x 10^23 CO 3 2- ions

Specify Mole of oxygen atoms means 6.02 x 10^23 O atoms Mole of oxygen molecules means 6.02 x 10^23 O 2 molecules (2 x 6.02 x 10^23 O atoms)

Molar mass: mass of one mole of the substance. Carbon-12 has a molar mass of exactly 12g/mol Ethanol, molecular formula C 2 H 5 OH MW = 46.1 amu Molar mass = 46.1 g/mol For all substances, the molar mass in grams per mole is numerically equal to the formula weight in atomic mass units.

MOLE CALCULATIONS -Conversion from mol to g -Ethanol is 46.1 g/mol 1 mol C 2 H 5 OH = 46.1 g Cr 2 H 5 OH To convert grams to moles

You need to prepare acetic acid from 10.0g of ethanol, C 2 H 5 OH. Convert 10.0g C 2 H 5 OH to moles C 2 H 5 OH by multiplying by the appropriate conversion factor.

3.3 MASS PERCENTAGES FROM THE FORMULA -Mass percentage of A as the parts of A per hundred parts of the total, by mass.

3.4 ELEMENTAL ANALYSIS: PERCENTAGES OF CARBON, HYDROGEN, AND OXYGEN -To determine the formula of a new compound is to obtain its percentage composition. -Ex. Burn a sample of the compound of known mass and get CO 2 and H 2 O. -Next relate the masses of CO 2 and H 2 O to the masses of carbon and hydrogen. -Then calculate the mass percentages of C and H. -Determine the mass percentage of O by difference.

3.5 DETERMINING FORMULAS -Empirical formula (simplest formula) – formula of a substance written with the smallest integer subscripts. -For most ionic substances, the empirical formula is the formula of the compound. -This is often not the case for molecular substances. -Hydrogen peroxide: molecular formula is H 2 O 2 -The empirical formula (just tells you the ratio of numbers of atoms in the compound) HO

Why do ionic compounds usually have the same formula and empirical formula? No multiple proportions like molecules.

Compounds with different molecular formulas can have the same empirical formula. They will also have the same percentage composition. Acetylene C 2 H 2 Benzene C 6 H 6 Same empirical formula CO, different molecular formulas, different chemical structures.

To obtain the molecular formula of a substance, you need to know: 1.The percentage composition, from which the empirical formula can be determined 2.Molecular weight

EMPIRICAL FORMULA FROM THE COMPOSITION -You can find the empirical formula from the composition of the compound by converting masses of the elements to moles.

Ex. 3.10

MOLECULAR FORMULA FROM EMPIRICAL FORMULA -Molecular formula of a compound is a multiple of its empirical formula. -Acetylene C2H2 is equivalent to (CH)2 -Benzene C6H6 is equivalent to (CH)6 Molecular weight = n x empirical formula weight

N = number of empirical formula units in the molecule. Molecular formula is obtained by multiplying the subscripts of the empirical formula by n n = molecular weight/empirical formula weight

After you find the empirical formula, calculate its empirical formula weight. From an experimental determination of its molecular weight, you can calculate n and then the molecular formula.

Ex. 3.11

STOICHIOMETRY: QUANTITATIVE RELATIONS IN CHEMICAL REACTIONS Stoichiometry – the calculation of the quantities of reactants and products involved in a chemical reaction. Relationship between mass and moles.

3.6 MOLAR INTERPRETATION OF A CHEMICAL EQUATION N 2 + 3H 2 = 2NH 3 How much hydrogen is required to give a particular quantity of ammonia? One N 2 One mole of N 2 reacts with Three H 2 three moles of H 2 to give two Two NH 3 moles of NH 3.

Write pg. 81

3.7 AMOUNTS OF SUBSTANCES IN A CHEMICAL REACTION -Balanced chemical equation relates the amounts of substances in a reaction. -The coefficients can be given a molar interpretation and can be used to calculate the moles of product obtained from any given moles of reactant.

Ex. You have a mixture of H 2 and N 2 and 4.8 mol H 2 in this mixture reacts with N 2 to produce NH 3. How many moles of NH 3 can you produce from this quantity of H 2 ? Multiplying any quantity of H 2 by this conversion factor converts that quantity of H 2 to the quantity of NH 3 as specified by the balanced chemical equation.

To calculate the quantity of NH 3 produced from 4.8 mol H 2, you write 4.8 mol H 2 and multiply this by: 4.8 mol H2 x 2 mol NH3/3 mol H2 = 3.2 mol NH3

Calculate the moles of reactant needed to obtain the specified moles of product. Setting up the conversion factor: refer to balanced equation and place the quantity you are converting from on the bottom and the quantity you are converting to on the top 3 mol H2 / 2 mol NH3 Converts from mol NH3 to mol H2

Practice problem: how much hydrogen is needed to yield 907 kg of ammonia? The balanced chemical equation directly relates moles of substances, not masses. *write conversion

Grams of A x conversion factor: g A to mol A x conversion factor: mol A to mol B x conversion factor: mol B to g B = grams of B

Example 3.12

3.8 LIMITING REACTANT; THEORETICAL AND PERCENTAGE YIELDS -Limiting reactant (reagent) – the reactant that is entirely consumed when a reaction goes to completion. -A reactant that is not completely consumed is often referred to as an excess reactant. -Once one of the reactants is used up, the reaction stops.

Analogy: your plant has in stock 300 steering wheels and 900 tires, plus an excess of every other needed component. How many autos can you assemble from this stock? 1 steering wheel + 4 tires + other components = 1 auto One way to solve this problem is to calculate the number of autos that you could assemble from each component. 300 steering wheels = 300 autos, 900 tires = 225 autos Tires are the limiting reactant.

Now consider a chemical reaction, the burning of hydrogen in oxygen. 2H2 + O2 = 2H2O H must be the limiting reactant. O is the excess reactant.

Ex 3.13

Ex 3.14

Theoretical yield: maximum amount of product that can be obtained by a reaction from given amounts of reactants. Actual yield of a product may be much less for several possible reasons. 1. Some product may be lost during the process of separating it from the final reaction mixture. 2. There may be other competing reactions that occur simultaneously with the reactant. 3. Many reactions appear to stop before they reach completion; they give mixtures of reactants and products.

Percentage yield of product is the actual yield expressed as a percentage of the the theoretical yield. Percentage yield = (actual yield/theoretical yield) x 100%

Ex. Acetic acid